Understand pointwise vs uniform convergence and their implications
When we have a sequence of functions, we can ask: does the sequence converge? Pointwise convergence requires fₙ(x) → f(x) for each x separately. Uniform convergence is stronger: all points converge at the same rate.
This distinction matters enormously! Uniform convergence preserves continuity and allows us to interchange limits with integrals. We'll see examples where pointwise convergence fails to preserve these properties.
A sequence fₙ converges pointwise to f if for each fixed x, the sequence of numbers fₙ(x) converges to f(x). The rate can vary dramatically across different points.
Converges pointwise on [0,1] but not uniformly
For ε = 0.10, need N ≥ 4
A sequence of functions fₙ converges pointwise to f if for each x:
∀ε > 0, ∃N such that n ≥ N ⟹ |fₙ(x) - f(x)| < ε
Note: The N can depend on both ε and x. Different points may converge at different rates!
Uniform convergence requires a single N that works for all points simultaneously. The entire graph of fₙ must eventually fit inside an ε-tube around the limit function.
sup|fₙ(x) - f(x)| does not → 0; no uniform N exists
fₙ → f uniformly if the same N works for all x simultaneously:
∀ε > 0, ∃N such that n ≥ N ⟹ sup|fₙ(x) - f(x)| < ε
Equivalently: ‖fₙ - f‖∞ → 0 as n → ∞.
Key insight: Uniform convergence means the entire graph of fₙ eventually fits inside an ε-tube around the limit function.
Uniform convergence preserves continuity: if each fₙ is continuous and fₙ → f uniformly, then f is continuous. Pointwise convergence does NOT guarantee this!
Each fₙ is continuous, but the limit has a jump discontinuity at x=1. Pointwise convergence does NOT preserve continuity.
Theorem: If fₙ → f uniformly and each fₙ is continuous, then f is continuous.
The converse is false: pointwise convergence of continuous functions can yield a discontinuous limit (as seen with fₙ(x) = xⁿ).
Proof idea: Given ε, find N so ‖fₙ - f‖∞ < ε/3, then use continuity of fₙ.
A powerful tool for proving uniform convergence of series. If |aₙ(x)| ≤ Mₙ for all x and Σ Mₙ converges, then Σ aₙ(x) converges uniformly.
Uniformly convergent since |cos(nx)/n²| ≤ 1/n² and Σ1/n² converges
Since Σ Mₙ converges, the M-test guarantees uniform convergence of Σ aₙ(x).
Theorem: Let Σ aₙ(x) be a series of functions on a domain D. If there exist constants Mₙ such that:
Then Σ aₙ(x) converges uniformly (and absolutely) on D.
Key idea: The Mₙ bound the functions uniformly, letting us control convergence independently of x.
When can we swap the order of limit and integral? Uniform convergence guarantees lim ∫fₙ = ∫ lim fₙ, but pointwise convergence can fail spectacularly.
lim ∫fₙ dx
0.4000
∫(lim fₙ) dx
0.0000
Uniform convergence ⟹ we can interchange limit and integral. lim ∫fₙ = ∫f = 0.
Theorem: If fₙ → f uniformly on [a, b], then:
limn→∞ ∫ab fₙ(x) dx = ∫ab f(x) dx
Warning: Pointwise convergence is NOT enough! The counterexamples above show that ∫fₙ can stay constant even as fₙ → 0.
If fₙ → f uniformly and each fₙ is continuous, then f is continuous. (Pointwise convergence does NOT suffice!)
If fₙ → f uniformly on [a, b], then lim ∫fₙ = ∫f. We can pass the limit under the integral sign.
If |aₙ(x)| ≤ Mₙ and Σ Mₙ converges, then Σ aₙ(x) converges uniformly.
If fₙ' → g uniformly and fₙ(x₀) converges for some x₀, then (lim fₙ)' = g. Derivatives require stronger conditions.